Question

How to prove that
(x^2)-(33y^3)=10[mod11] has no solutions?

Answer 1

I'm assuming x is an integer? Then maybe you can maybe attempt a direct proof using cases, (case 1 being even integers and case 2 being odd integers)

Answer 2

**x and y are integers

Answer 3

yes sorry,\[for x,y \in Z\]

Answer 4

Oh okay, I'll give that a go, thanks :)

Answer 5

i may be confused because it is past my bedtime, but isn't this the same as asking to prove that
\[x^2\equiv 10 (11)\] as no solutions?

Answer 6

on account of \(33y^3\equiv 0(11)\)

Answer 7

Was the original conjecture meant to say:

x^2 - 33y^2 ≡ 10mod11 ? instead of equals?

Answer 8

"equals" means equivalent working modulo an integer
\(33n\equiv 0(11)\) for any \(n\) so either this is straightforward or (as is always possible) i am confused

Answer 9

Sorry, the full question is:
Show that the Diophantine equation \[x ^{2} - 33y ^{3}=10\ x,y \in Z\].
has no solutions
[Hint: Consider the equation modulo 11.]

Answer 10

ok so once you decide it has no solutions modulo 11, then what?

Answer 11

ooh i think i see what is going on

Answer 12

i must be missing something, because i think the hint is more a hindrance than a help

Answer 13

prove \[x ^{2} - 33y ^{3}=10\ x,y \in Z\] has no integer solutions
if \[x ^{2} - 33y ^{3}=10\ x,y \in Z\] then
\[x ^{2}=10 +33y^3\ x,y \in Z\] meaning
\[x^2\equiv 10 \text{mod}(11)\] which his impossible

Answer 14

now the question makes sense to me
i didn't understand the original one

Answer 15

proving that you cannot solve \(x^2\equiv 10 (\text{ mod } 11)\) should be okay right?

Answer 16

Oh I see what you mean now! Yes I think I'm okay now! I was interpreting it incorrectly I think

Thankyou!